well, AO and OB are both radii, and therefore, both of them are 10 inches long, meaning the triangle has a height of 10 and a base of 10.
let's also bear in mind that the central angle of that segment as well as sector is 90°.
[tex]\bf \textit{area of a sector or a circle}\\\\ A=\cfrac{\theta \pi r^2}{360}~~ \begin{cases} r=radius\\ \theta =angle~in\\ \qquad degrees\\[-0.5em] \hrulefill\\ \theta =90\\ r=10 \end{cases}\implies A=\cfrac{(90)\pi (10)^2}{360}\implies A=25\pi \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{sector's area}}{25\pi }~~-~~\stackrel{\textit{triangle's area}}{\cfrac{1}{2}(10)(10)}\implies 25\pi -50\implies \stackrel{\textit{area of the segment}}{\approx~~28.54}[/tex]
Answer:
28.5
Step-by-step explanation:
other comment i just rounded it
